Not a problem! The "fixed point" refers to the value(s) of \(f_{n-1}\) that when you work out \(f_n\), you get the same value that \(f_{n-1}\) was, this putting you in a never ending loop and hence is a fixed point (or equilibrium).

Here's the method for finding the fixed point (explained pretty loosely):

-> By definition the fixed point means that the next number should be the same as the current one which is equivalent to saying \(...=f_{n-2}=f_{n-1}=f_n=f_{n+1}=...\) This means we can forget about the subscripts and treat everything as the same unknown \(f\).

-> This means at the fixed point the equation \(f_n=3f_{n-1}-60\) transforms into \(f=3f-60\).

-> Now you just solve for \(f\) and that will be your fixed point

In this example, \(f = 30\) and substituting it into the original difference equation we get:

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